Question

I'm really having troubles with this question, could someone please help?!

A ballplayer catches a ball 3.0s after throwing it vertically upward. With what speed did he throw it, and what height did it reach? Gravity is 9.8m/s^2

Answer 1

Thats a hard question ..

Answer 2

I know right! I don't believe there is enough information to answer the question. The back of the book says that the speed is 15m/s and the height is 11m. I can't figure out how they came up with that answer

Answer 3

I figured it out... you need to divide the time in half so it will equal 1.5 so that's how long it took it to ascend and then plug the numbers in to the \[V = V _{0} + at\] equations. That comes out to be 14.7m/s. But when using significant figures it rounds to 15m/s.

Answer 4

its a tricky question, lets see the scenario when the ball is at its highest position, and we know that its velocity will be zero there, and total time is given 3 seconds so body will take the same time on reaching to the hands as it takes to reach to the highest point, so 3/2=1.5seconds ball will take to reach to the hands from its highest point, and we know that acceleration here will be due to gravity i.e 9.8m/s^2, now we will form data to make work easy ,
velocity of ball at its highest point=vi=o
time taken to reach to hands from highest point=t=1.5 seconds
velocity at reaching hands=vf=?
a=g=9.8m/s^2\[vf=vi+at\] \[vf=0+2\times9.8\] \[vf=14.7=15m/s\] now for height \[vf^2-vi^2=2aS\] \[15^2-0^2=2\times9.8\times S\] \[S=11.47 metres\]